Past exam of the theoretical physics major option course of the University of Oxford / 2022 / 1

For the sawtooth Ising chain, there is a source inconsistency: the printed Ising spin values are $\{-1,r+1\}$, with no definition of $r$. The requested transfer matrix follows for $r=0$, namely the usual Ising model with values $\{-1,+1\}$. For arbitrary $r$, summing an apex spin actually givesFor example, $r=1$, $s=s^{\prime }=-1$ and $\beta \bar{J}=1$ give the bracket $e^{-2}+e^4$, whereas the claimed hyperbolic cosine gives $e^{-2}+e^2$. Thus the printed identity is false literally. The rest of Question 1 uses the intended symmetric Ising spin values, $r=0$.

For a fixed configuration of base spins, each apex spin appears in only one local Boltzmann factor. The apex spins are therefore conditionally independent, and their finite sums factorize:Only the base spins carry information from one triangle to the next. The apex spins have already been summed out, so no apex index remains in the transfer matrix; the final trace implements the periodic boundary condition.

In the $s=+1,-1$ basis the transfer matrix and its eigenvalues areFor finite positive temperature, $a,b>0$ and ${\lambda }_{+}>|{\lambda }_{-}|$. Consequently $Z={\lambda }_{+}^L+{\lambda }_{-}^L$ and $L^{-1}\log Z\to \log {\lambda }_{+}$ in the thermodynamic limit. There are $2L$ spins, so the Helmholtz free energy per spin isThe factor $1/2$ is essential: one transfer matrix step counts two spins. For completeness, if the undefined $r$ is retained, set $a_r=A_{-1,-1}^{(r)}$, $d_r=A_{r+1,r+1}^{(r)}$ and $b_r=A_{-1,r+1}^{(r)}$. The same thermodynamic limit gives $f_r=-(2\beta {)}^{-1}\log [(a_r+d_r+\sqrt{(a_r-d_r{)}^2+4b_r^2})/2]$ whenever these are two distinct spin values.

Differentiate the Helmholtz free energy at fixed couplings. With $\lambda ={\lambda }_{+}$ from solution, the entropy per spin isIt is particularly useful to writeAt $\bar{J}=J$, this is $3e^{\beta J}+e^{-3\beta J}$, so $f=-J/2-(\log 3)/(2\beta )+o({\beta }^{-1})$. For fixed $\bar{J}>J$, the first exponential dominates alone, giving $f=(J-2\bar{J})/2+o({\beta }^{-1})$. Hence the residual entropy isThe finite positive difference $\bar{J}-J$ must be kept fixed in the second limit; taking it to zero first gives the other answer.

At high temperature, $\lambda \to 4$ and $\beta {\partial }_{\beta }\log \lambda \to 0$, whenceAll $2^{2L}$ configurations become equally probable because their Boltzmann factors tend to one. The two available states per spin, rather than the coupling values, determine this entropy.

For one triangle, equal base spins $s=s^{\prime }=\sigma$ give energy $J+2\bar{J}\sigma \bar{s}$. Their minimum occurs at $\bar{s}=-\sigma$ and is $J-2\bar{J}$. Opposite base spins give energy $-J$, independently of the apex spin. The all-equal configurations have energy $J+2\bar{J}$.

When $\bar{J}=J$, the six configurations with precisely one spin different from the other two all have minimum energy $-J$. For a fixed first base spin, there is one minimum choice if the next base spin is the same, and two choices of apex if it is opposite. The ground-configuration counting transfer matrix is thereforeThis ground-state counting of the frustrated sawtooth Ising chain proves the extensive geometric frustration and the residual entropy $k_B{\lim }_{L\to \infty }\log {\Omega }_L/(2L)=(k_B/2)\log 3$.

When $\bar{J}>J$, $J-2\bar{J}<-J$. Every triangle must have equal base spins and the opposite apex spin. Shared base spins propagate that choice around the chain, leaving just two global ground states, of energy $L(J-2\bar{J})$. Their entropy per spin is $k_B\log 2/(2L)\to 0$.

For couplings favoring ferromagnetism, write $J=-K$, $\bar{J}=-\bar{K}$ with $K,\bar{K}>0$. Both orientations of an all-equal triangle minimize every bond simultaneously. The common base spins then make the whole chain uniform. There are two global ground states for either $K=\bar{K}$ or $K\ne \bar{K}$, so no corresponding distinction in residual entropy occurs. This statement assumes all bonds remain nonzero.